GPT 6 Astra Ultra attempt on Math-6253
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Contents
Erdős Problem 1199 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
For every map $c:\mathbb N\to\{0,1\}$, is there an infinite set $A\subseteq\mathbb N$ and $i\in\{0,1\}$ such that \[c(a+b)=i\qquad\text{for all }a,b\in A?\] The requirement includes $a=b$, so all doubled elements $2a$ are included.CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.ATTACK PLAN AND WORK
Infinite Ramsey extraction reduces the difficulty to a precise mismatch. First choose an infinite set $X$ on which the doubled values $c(2x)$ have one color $d$, by pigeonholing. Color each unordered pair of distinct members of $X$ by $c(x+y)$. The nested-tail proof of infinite Ramsey's theorem yields an infinite $A\subseteq X$ whose distinct-pair sums all have one color $e$. If $d=e$, this $A$ solves the full problem, including repeated summands.The mismatch $d\ne e$ cannot simply be ruled out on the extracted set. For the coloring $c(n)=v_2(n)\bmod2$ and $A=\{2^{2j}:j\ge1\}$, distinct-pair sums have even 2-adic valuation, while doubled elements have odd valuation. This example refutes the proposed inference from off-diagonal homogeneity; it is not a counterexample to the original existence question for that coloring.
The full conclusion does hold for every eventually periodic coloring. If the period is $q$ beyond $N$, take all sufficiently large positive multiples of $q$ as $A$. Every sum, including $2a$, lies in the same residue class zero modulo $q$ beyond $N$, and hence has the same color.